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Is y < 0 (1) x + y = 1 (2) x^2 < y^2

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Is y < 0 (1) x + y = 1 (2) x^2 < y^2  [#permalink]

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New post 10 Jun 2018, 04:42
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

42% (01:46) correct 58% (01:17) wrong based on 31 sessions

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Re: Is y < 0 (1) x + y = 1 (2) x^2 < y^2  [#permalink]

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New post 10 Jun 2018, 06:53
1
Well, firstly neither statement is sufficient. Testing numbers is enough to figure this out.

(1) Let y = -1 (so y < 0), x = 2. Let y = 2 (so y > 0), x = -1. Since with (1) y can be less than 0 or not, it is not sufficient.

(2) Let y = -3 (so y < 0), then y^2 = 9. So long as x^2 can be <9 then y = -3 is valid. We can have x be 1 or 2 or whatever and it'd be valid. Now, let y = 3 (so y > 0), then y^2 = 9. Same case as above. Since y can be less than 0 or not, (2) is also not sufficient.

Together? Well let's re-write the equation in statement (1): x + y = 1 <=> x = 1 - y.

Thus with statement (2), we have (1-y)^2 < y^2. I can't solve this normally as I haven't practiced inequalities in a long time, so let's plug in numbers XD. Let's try y = 0. In this case, we have 1 < 0, which is impossible. Let's try y = 1, in which case we have 0 < 1. Let's try a number larger than 1. Say 2. We then have 1 < 4, which is fine. In fact, any positive number >= 1 will have the effect, because then the absolute value of 1-y will always be 1 less than y, and thus (1-y)^2 will always be < y^2. Now, what about y < 0? Say -1? We then have 4 < 1, which is impossible. We can try a much larger negative number, like -100. We'd have 101^2 < 100^2, again is not possible. We have something similar to the previous case, except that the absolute value of 1-y will always be 1 higher than y, which is impossible due to the inequality. If we tested fractions too, we'd figure out that some of the fractions are not possible, but it is unnecessary to test them as we only care if y is < 0 or not. y may also not be < than 1/2 for example, but that still fits if we determine that y must be > 0, which in this case we do.

Thus, answer is C.
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Not a professional entity or a quant/verbal expert or anything. So take my answers with a grain of salt.

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Re: Is y < 0 (1) x + y = 1 (2) x^2 < y^2  [#permalink]

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New post 23 Jul 2018, 01:32
1
Bunuel wrote:
Is y < 0

(1) x + y = 1
(2) x^2 < y^2


Clearly 1 and 2 both are individually insufficient

lets try for option C
rearranging as we know that square terms are positive
we get,
X^2 - Y^2 < 0
(X+Y)(X-Y) < 0
But X + Y = 1 (From statement 1)
So X-Y < 0
X < Y
So for X < Y statement 1 will never be true if Y<0
Hence sufficient, Y is not < 0

ANSWER : C
Re: Is y < 0 (1) x + y = 1 (2) x^2 < y^2 &nbs [#permalink] 23 Jul 2018, 01:32
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